The angle of elevation of a …

Height of jet plane {tex}= 1500 \sqrt { 3 } \mathrm { m }{/tex}
PQ = y and PR = x
where Q is right below A (1^st position)
and R is right below B (2^nd position)
Using {tex}\triangle PRB{/tex} ,{tex}\frac { x } { 1500 \cdot \sqrt { 3 } } = \cot 30 ^ { \circ }{/tex}  
{tex}\Rightarrow \frac { x } { 1500 \cdot \sqrt { 3 } } = \cot 30 ^ { \circ }{/tex}

{tex}\Rightarrow \frac { x } { 1500 \sqrt { 3 } } = \sqrt { 3 }{/tex}
{tex}\Rightarrow x = 1500 \cdot \sqrt { 3 } \times \sqrt { 3 }{/tex}
{tex}\therefore{/tex} x = 4500 m ........(i)
Using {tex}\triangle PQA{/tex}, {tex}\frac { y } { 1500 \sqrt { 3 } } = \cot 60 ^ { \circ }{/tex}
{tex}\Rightarrow \frac { \mathrm { y } } { 1500 \sqrt { 3 } } = \frac { 1 } { \sqrt { 3 } }{/tex}
{tex}y = \frac { 1500 \sqrt { 3 } } { \sqrt { 3 } } = 1500 m{/tex}......(ii)
Subtracting (ii) from (i), we get x - y = 4500 - 1500 = 3000 m
Distance traveled in 15 seconds = 3000 m
Distance traveled in 1 second = {tex}\frac { 3000 } { 15 } = 200 \mathrm { m } / \mathrm { sec }{/tex}
Hence, speed of jet plane = {tex}\frac { 200 } { 1000 } \times 60 \times 60 \mathrm { km } / \mathrm { hr } = 720 \mathrm { km } / \mathrm { hr }{/tex}